Permutations and Combinations — H2 Maths help

Permutations and combinations underpin probability and a wide range of counting problems in H2 Mathematics. The central skill is deciding whether order matters: if it does, you are counting permutations; if it does not, you are counting combinations.

What you need to know

Permutations (order matters): the number of ways to arrange $r$ objects chosen from $n$ distinct objects in a row is ${}^{n}P_r = \dfrac{n!}{(n-r)!}$ . Arranging all $n$ objects gives $n!$ ways.
Combinations (order does not matter): the number of ways to choose $r$ objects from $n$ distinct objects is ${}^{n}C_r = \dfrac{n!}{r!\,(n-r)!}$ . Note ${}^{n}C_r = {}^{n}C_{n-r}$ .
Objects that must stay together: treat the group as a single block. Count arrangements of the blocks, then multiply by the internal arrangements of the block.
Objects that must be separated (gaps method): first arrange the remaining objects, then place the restricted objects in the gaps between them (including the ends).
Circular arrangements: fix one object to remove rotational equivalence; the remaining $n-1$ objects are arranged in $(n-1)!$ ways.
Selections with conditions: use ${}^{n}C_r$ for each case (must-include, must-exclude, etc.) and add the valid cases.

Worked example 1 — Row arrangement with a restriction (people must be together)

Problem. Eight students — Alice, Bob, and six others — are to be arranged in a row. Find the number of arrangements in which Alice and Bob are next to each other.

Solution.

Treat Alice and Bob as a single block. There are now $7$ objects to arrange (the block plus the 6 others):

7! = 5040 \text{ ways to arrange the 7 objects.}

Within the block, Alice and Bob can swap places:

2! = 2 \text{ internal arrangements.}

Total arrangements with Alice and Bob together:

7! \times 2! = 5040 \times 2 = 10\,080.

Answer: $10\,080$ arrangements.

Worked example 2 — Row arrangement with a restriction (people must be apart)

Problem. Using the same 8 students, find the number of arrangements in which Alice and Bob are not next to each other.

Solution.

Use complementary counting:

\text{Total arrangements} - \text{arrangements with Alice and Bob together.}

Total arrangements of 8 students:

8! = 40\,320.

From Worked example 1, arrangements with Alice and Bob together $= 10\,080$ .

Therefore, arrangements with Alice and Bob not together:

40\,320 - 10\,080 = 30\,240.

Answer: $30\,240$ arrangements.

Worked example 3 — Selection (committee with conditions)

Problem. A committee of 4 is to be chosen from 5 men and 4 women. Find the number of ways to form the committee if it must contain at least 2 women.

Solution.

The cases satisfying “at least 2 women” are exactly 2, 3, or 4 women.

Case 1: exactly 2 women, 2 men.

{}^{4}C_2 \times {}^{5}C_2 = 6 \times 10 = 60.

Case 2: exactly 3 women, 1 man.

{}^{4}C_3 \times {}^{5}C_1 = 4 \times 5 = 20.

Case 3: exactly 4 women, 0 men.

{}^{4}C_4 \times {}^{5}C_0 = 1 \times 1 = 1.

Total:

60 + 20 + 1 = 81.

Answer: $81$ committees.

Common mistakes

Confusing permutations with combinations. Ask: does the order of selection matter? Choosing a team of 4 does not depend on the order chosen ( ${}^{n}C_r$ ); awarding gold/silver/bronze/fourth place from 4 finalists does ( ${}^{n}P_r$ ).
Forgetting internal arrangements in the block method. After treating a group as a block, always multiply by the number of ways to arrange the objects inside the block.
Not accounting for both ends in the gaps method. If 6 objects are arranged in a row, there are 7 gaps (5 between objects plus 2 ends); failing to count the end gaps leads to an undercount.
Double-counting in circular arrangements. Do not apply $(n-1)!$ and then also divide by a reflection symmetry unless the problem explicitly states that clockwise and anticlockwise are indistinguishable (e.g., a necklace); for a round table, $(n-1)!$ is the standard result.